Question

$2NO\left(g\right)+C{l}_2\left(g\right)\to 2NOCl\left(g\right)$
The following data were collected. All the measuremnets were taken at 263 K.

Experiment No.	Initial [NO] (M)	Initial $\left[C_2\right]$ (M)	Initial rate of disappearance of $C{l}_2$ (M/min)
1	0.15	0.15	0.60
2	0.15	0.30	1.20
3	0.30	0.15	2.40
4	0.25	0.25	?

(a) Write the expression for rate law.
(b) Calculate the value of rate constant and specify its units.
(c) What is the initial rate of disappearance of $C{l}_2$ in exp. 4?

Answer 1

(a) Rate law may be written as

Rate $=k\left[NO{]}^p\right[C{l}_2{]}^p$

The initial rate becomes

$(Rate{)}_0=k[NO{]}^P[C{l}_2{]}^p$

Comparing experiment 1 and 2

$(Rate{)}_1=k(0.15{)}^p(0.15{)}^p=0.60$ ...(i)

$(Rate{)}_2=k(0.15{)}^p(0.30{)}^p=1.20$ ...(ii)

Dividing equation (ii) by (i)

$\frac{(Rate{)}_2}{(Rate{)}_2}=\frac{k\left(0.15{)}^p\right(0.30{)}^p}{k\left(0.15{)}^p\right(0.1{)}^p}=\frac{1.20}{0.60}$

or $2^q=2^1$

$q=1$

order with respect to $C{l}_2=1$

Comparing experiment 1 and 3

$(Rate{)}_1=k(0.15{)}^p(0.15{)}^q=0.60$ ...(i)

$(Rate{)}_3=k(0.30{)}^p(0.15{)}^q=2.40$ ...(ii)

Dividing equation (ii) by (i)

$\frac{(Rate{)}_3}{(Rate{)}_1}=\frac{k\left(0.30{)}^p\right(0.15{)}^q}{k\left(0.15{)}^p\right(0.15{)}^q}=\frac{2.40}{0.60}$

or $2^r=4$

$2^q=2^2$

$p=2$

Thus order with respect to NO is 2

Rate law $=k\left[NO{]}^2\right[C{l}_2{]}^1$

(b) The rate law for the reaction Rate $=k\left[NO{]}^2\right[C{l}_2]$

Rate constant can be calculated by substituting the value of rate, [NO] and $\left[C{l}_2\right]$ for any experiment

$k=\frac{Rate}{\left[NO{]}^2\right[C{l}_2]}=\frac{0.60}{\left(0.15{)}^2\right(0.15)}$

$=\frac{0.60}{0.003375}$

$=177.77mo{l}^{-2}{L}^{2}mi{n}^{-1}$

(c) Let initial rate of disappearance of $C{l}_2$ in exp. 4 is $r_4$

$\therefore r_4=k\left[NO{]}^2\right[C{l}_2]$

$=177.77\times \left(0.25{)}^2\right(0.25)$